Determine the probability that for a positive integer N chosen at random between 1000 and 9999 inclusively, the sum of the products of pairs of digits in N is equal to the sum of products of triplets of its digits.

Only a few combinations of digits will result in the sum of the products of its pairs and triplets being equal (keeping in mind the first digit of each number is a non-zero digit):

The sets of digits are... {0,0,0,x} : x is a non-zero decimal digit, {0,2,3,6}, {0,2,4,4} and {0,3,3,3}, {1,1,2,3}.

0, 0, 0, x --->  9
1000, 2000, 3000, 4000, 5000, 6000, 7000, 8000, 9000

0, 2, 3, 6 ---> 18
2036, 2063, 2306, 2360, 2603, 2630, 3026, 3062, 3206, 3260, 3602, 3620, 6023, 6032, 6203, 6230, 6302, 6320

0, 2, 4, 4 --->  9
2044, 2404, 2440, 4024, 4042, 4204, 4240, 4402, 4420

0, 3, 3, 3 --->  3
3033, 3303, 3330

1, 1, 2, 3 ---> 12
1123, 1132, 1213, 1231, 1312, 1321, 2113, 2131, 2311, 3112, 3121, 3211

The probablility is then (9+18+9+3+12)/(9999-1000+1) = 51/9000 = 0.00566

Edited on November 8, 2009, 9:06 pm

Posted by Dej Mar on 2009-11-08 21:03:18